More simple and powerful is to devide nunerator and denominator by n obtaining...
(1)
... so that is...
(2)
Kind regards
Is my working of this problem ok?
f:N->R
f(n) =
prove f(n) = O(n) < - - - thats big O with a "-" in it
= since each term is >=0
<= since log2n<=n
<= since n>=1
=
=
=3n
since each term >= 0
>= since log2n is >= 0 for n>=1
=
=n
Therfore
f(n) = O(n)
I really hope i am doing this correctly
Thanks for any help and input.
thanks for the help Chisigma, but what is more simple to you seems a step beyond what I am learning at the moment (I think).
I just tried to answer the problem as close to the example solutions I had studied. We didn't divide the numerator, just worked it out in the style I have above.
I am VERY new to discrete maths, is my attempt at the solution completely incorrect then? or even close?
Hi, Can someone look at my original post, (even though I spelled "upper" incorrectly").
Chisigma gave me an answer I don't really understand. She said " a more simple and powerful...." but I don't know if this implies my attempt was ok, but could be better done, or if my attempt was wrong.
But I need to know if my attempt at the solution was way off, or good enough.
Thanks for any help/input